例子:
absolute="/foo/bar"
current="/foo/baz/foo"
# Magic
relative="../../bar"
我如何创造魔法(希望不是太复杂的代码…)?
例子:
absolute="/foo/bar"
current="/foo/baz/foo"
# Magic
relative="../../bar"
我如何创造魔法(希望不是太复杂的代码…)?
当前回答
我把你的问题作为一个挑战,用“可移植的”shell代码来编写它,即。
考虑到POSIX外壳 没有数组之类的bashisms 避免像打瘟疫一样打外部电话。脚本中没有一个分叉!这使得它非常快,特别是在有显著分叉开销的系统上,比如cygwin。 必须处理路径名中的glob字符(*,?,[,])
它运行在任何POSIX兼容shell (zsh, bash, ksh, ash, busybox,…)上。它甚至包含一个测试套件来验证其操作。路径名的规范化留作练习。: -)
#!/bin/sh
# Find common parent directory path for a pair of paths.
# Call with two pathnames as args, e.g.
# commondirpart foo/bar foo/baz/bat -> result="foo/"
# The result is either empty or ends with "/".
commondirpart () {
result=""
while test ${#1} -gt 0 -a ${#2} -gt 0; do
if test "${1%${1#?}}" != "${2%${2#?}}"; then # First characters the same?
break # No, we're done comparing.
fi
result="$result${1%${1#?}}" # Yes, append to result.
set -- "${1#?}" "${2#?}" # Chop first char off both strings.
done
case "$result" in
(""|*/) ;;
(*) result="${result%/*}/";;
esac
}
# Turn foo/bar/baz into ../../..
#
dir2dotdot () {
OLDIFS="$IFS" IFS="/" result=""
for dir in $1; do
result="$result../"
done
result="${result%/}"
IFS="$OLDIFS"
}
# Call with FROM TO args.
relativepath () {
case "$1" in
(*//*|*/./*|*/../*|*?/|*/.|*/..)
printf '%s\n' "'$1' not canonical"; exit 1;;
(/*)
from="${1#?}";;
(*)
printf '%s\n' "'$1' not absolute"; exit 1;;
esac
case "$2" in
(*//*|*/./*|*/../*|*?/|*/.|*/..)
printf '%s\n' "'$2' not canonical"; exit 1;;
(/*)
to="${2#?}";;
(*)
printf '%s\n' "'$2' not absolute"; exit 1;;
esac
case "$to" in
("$from") # Identical directories.
result=".";;
("$from"/*) # From /x to /x/foo/bar -> foo/bar
result="${to##$from/}";;
("") # From /foo/bar to / -> ../..
dir2dotdot "$from";;
(*)
case "$from" in
("$to"/*) # From /x/foo/bar to /x -> ../..
dir2dotdot "${from##$to/}";;
(*) # Everything else.
commondirpart "$from" "$to"
common="$result"
dir2dotdot "${from#$common}"
result="$result/${to#$common}"
esac
;;
esac
}
set -f # noglob
set -x
cat <<EOF |
/ / .
/- /- .
/? /? .
/?? /?? .
/??? /??? .
/?* /?* .
/* /* .
/* /** ../**
/* /*** ../***
/*.* /*.** ../*.**
/*.??? /*.?? ../*.??
/[] /[] .
/[a-z]* /[0-9]* ../[0-9]*
/foo /foo .
/foo / ..
/foo/bar / ../..
/foo/bar /foo ..
/foo/bar /foo/baz ../baz
/foo/bar /bar/foo ../../bar/foo
/foo/bar/baz /gnarf/blurfl/blubb ../../../gnarf/blurfl/blubb
/foo/bar/baz /gnarf ../../../gnarf
/foo/bar/baz /foo/baz ../../baz
/foo. /bar. ../bar.
EOF
while read FROM TO VIA; do
relativepath "$FROM" "$TO"
printf '%s\n' "FROM: $FROM" "TO: $TO" "VIA: $result"
if test "$result" != "$VIA"; then
printf '%s\n' "OOOPS! Expected '$VIA' but got '$result'"
fi
done
# vi: set tabstop=3 shiftwidth=3 expandtab fileformat=unix :
其他回答
我将只使用Perl来完成这个不那么简单的任务:
absolute="/foo/bar"
current="/foo/baz/foo"
# Perl is magic
relative=$(perl -MFile::Spec -e 'print File::Spec->abs2rel("'$absolute'","'$current'")')
test.sh:
#!/bin/bash
cd /home/ubuntu
touch blah
TEST=/home/ubuntu/.//blah
echo TEST=$TEST
TMP=$(readlink -e "$TEST")
echo TMP=$TMP
REL=${TMP#$(pwd)/}
echo REL=$REL
测试:
$ ./test.sh
TEST=/home/ubuntu/.//blah
TMP=/home/ubuntu/blah
REL=blah
这个答案并没有解决问题的Bash部分,但是因为我试图使用这个问题中的答案在Emacs中实现这个功能,所以我就把它扔到那里。
Emacs实际上有一个开箱即用的函数:
ELISP> (file-relative-name "/a/b/c" "/a/b/c")
"."
ELISP> (file-relative-name "/a/b/c" "/a/b")
"c"
ELISP> (file-relative-name "/a/b/c" "/c/b")
"../../a/b/c"
这个脚本只对路径名有效。它不需要任何文件存在。如果传递的路径不是绝对的,那么行为就有点不寻常,但是如果两条路径都是相对的,那么应该能正常工作。
我只在OS X上测试过,所以可能不太便携。
#!/bin/bash
set -e
declare SCRIPT_NAME="$(basename $0)"
function usage {
echo "Usage: $SCRIPT_NAME <base path> <target file>"
echo " Outputs <target file> relative to <base path>"
exit 1
}
if [ $# -lt 2 ]; then usage; fi
declare base=$1
declare target=$2
declare -a base_part=()
declare -a target_part=()
#Split path elements & canonicalize
OFS="$IFS"; IFS='/'
bpl=0;
for bp in $base; do
case "$bp" in
".");;
"..") let "bpl=$bpl-1" ;;
*) base_part[${bpl}]="$bp" ; let "bpl=$bpl+1";;
esac
done
tpl=0;
for tp in $target; do
case "$tp" in
".");;
"..") let "tpl=$tpl-1" ;;
*) target_part[${tpl}]="$tp" ; let "tpl=$tpl+1";;
esac
done
IFS="$OFS"
#Count common prefix
common=0
for (( i=0 ; i<$bpl ; i++ )); do
if [ "${base_part[$i]}" = "${target_part[$common]}" ] ; then
let "common=$common+1"
else
break
fi
done
#Compute number of directories up
let "updir=$bpl-$common" || updir=0 #if the expression is zero, 'let' fails
#trivial case (after canonical decomposition)
if [ $updir -eq 0 ]; then
echo .
exit
fi
#Print updirs
for (( i=0 ; i<$updir ; i++ )); do
echo -n ../
done
#Print remaining path
for (( i=$common ; i<$tpl ; i++ )); do
if [ $i -ne $common ]; then
echo -n "/"
fi
if [ "" != "${target_part[$i]}" ] ; then
echo -n "${target_part[$i]}"
fi
done
#One last newline
echo
这里的答案并不是每天都能用的。由于在纯bash中很难正确地做到这一点,我建议以下可靠的解决方案(类似于注释中的一个建议):
function relpath() {
python -c "import os,sys;print(os.path.relpath(*(sys.argv[1:])))" "$@";
}
然后,你可以得到基于当前目录的相对路径:
echo $(relpath somepath)
或者你可以指定路径相对于给定的目录:
echo $(relpath somepath /etc) # relative to /etc
一个缺点是这需要python,但是:
它在任何python >= 2.6中工作相同 它不要求文件或目录存在。 文件名可以包含更广泛的特殊字符。 例如,如果文件名包含 空格或其他特殊字符。 它是一个单行函数,不会使脚本混乱。
注意,包含basename或dirname的解决方案不一定更好,因为它们要求安装coreutils。如果有人有可靠而简单的纯bash解决方案(而不是令人费解的好奇心),我会感到惊讶。